Question

practice solving problems using factorials, permutations, and combinations. three students, angie, bradley, and carnell, are being selected for three student council offices: president, vice president, and treasurer. in each arrangement below, the first initial of each person’s name represents that person’s position, with president listed first, vice president second, and treasurer third. which shows the possible outcomes for the event? abc abc, bac, cba aaa, bbb, ccc abc, acb, bca, bac, cab, cba

Explanation:

We need to find the permutations of 3 students (Angie - A, Bradley - B, Carnell - C) for 3 distinct offices (president, vice - president, treasurer). The number of permutations of \(n\) distinct objects is \(n!\). For \(n = 3\), \(3!=3\times2\times1 = 6\). The permutations of A, B, C are ABC, ACB, BCA, BAC, CAB, CBA. Let's analyze each option:

• Option 1 (ABC): This is only one permutation, but we need all 6 permutations of 3 distinct people for 3 distinct positions.
• Option 2 (ABC, BAC, CBA): This is only 3 permutations, not all 6.
• Option 3 (AAA, BBB, CCC): These are not permutations of distinct people (repeating the same person for all positions, which is not possible as we have 3 distinct people for 3 distinct positions).
• Option 4 (ABC, ACB, BCA, BAC, CAB, CBA): This has all 6 permutations of A, B, C, which are the possible arrangements for 3 distinct people in 3 distinct offices.

Answer:

D. ABC, ACB, BCA, BAC, CAB, CBA